为何从引用转换为c_void指针需要双重转换？

Let's break down why that intermediate cast is necessary, and what type the compiler infers for _.

1. The Type Conversion Chain Explained #

First, let's map out the types step by step for your working code:

• bar is of type *mut foo
• &mut bar creates a mutable reference: &mut *mut foo
• The first cast as *mut *mut _ tells Rust to convert that mutable reference into a raw pointer matching the underlying nested type. The compiler infers _ as foo, so this cast produces *mut *mut foo
• The second cast as *mut *mut c_void converts *mut *mut foo to *mut *mut c_void—this is allowed because Rust explicitly permits converting raw pointers to/from *mut c_void (the FFI standard for "generic pointers").

2. Why Direct Conversion Fails #

When you skip the intermediate step and write &mut bar as *mut *mut c_void, Rust rejects it because its casting rules don't support jumping directly from a mutable reference (&mut *mut foo) to a raw pointer of a different nested pointer type (*mut *mut c_void).

Rust requires explicit, incremental steps here:

• First, convert the mutable reference to a raw pointer of the exact same nested type (*mut *mut foo). This is a safe, straightforward cast since &mut T can always be converted to *mut T.
• Then, convert that raw pointer to the target FFI-compatible type (*mut *mut c_void). Raw pointers to any type can be cast to pointers to c_void, so this step is valid.

3. What the *-ptr Error Message Means #

When you tried assigning &mut bar as *mut *mut _ to a u8, the compiler's *-ptr message is just a shorthand for "a raw pointer type"—specifically, it refers to *mut *mut foo. The compiler uses this generic term to highlight the type mismatch between the pointer and your expected u8, not because the type is unknown.

A More Explicit Alternative #

If you want to remove ambiguity, you can write the intermediate type explicitly instead of using _—it does the exact same thing:

struct foo;
let mut bar: *mut foo = std::ptr::null_mut();
unsafe {
    ffi(&mut bar as *mut *mut foo as *mut *mut std::os::raw::c_void);
}

内容的提问来源于stack exchange，提问作者Ryan C.